showing that $f_n(x)= x^n$ is no Cauchy Sequence

Question

Considering the space $C^0([0,1])$ of continuous functions on $[0,1]$, with the norm $||f|| = \max_{x \in [0,1]} |f(x)|$ I have to determine whether $f_n(x) = x^n, n \in \mathbb N$ is a cauchy sequence or not. My intuition tells me it is not. My question is now: How can I show that $$\left(\frac N m \right)^{\frac{N}{m-N}}-\left(\frac N m \right)^{\frac{m}{m-N}} \to 1$$ if $m \to \infty$?

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Some background

(In case you find an easier way.)

I am assuming it is a CS, so for $\varepsilon = 1/2$ I assume there is an $N$ such that $$||f_n - f_m|| < \epsilon \forall n,m \geq N$$, and now want to find a contradiction. Fixing $N=n$ I tried to show that we can make $||f_N-f_m|| = \max ( x^N - x^m)$ get arbitrarily close to 1 if we choose $m$ big enough. For that I determined the maximum of $x^N-x^m$ is at $x = \left(\frac n m \right)^{\frac{1}{m-n}}$ via setting the derivative to zero. Plugging this back into the expression of the norm, I got to that question.

Answer 1

A computational (low-tech) answer:

$$\begin{split} \left(\frac Nm\right)^{\frac{N}{m-N}} &= \left(\frac N{(m-N)+N}\right)^{\frac{N}{m-N}}\\ &=\left(\frac 1{\frac{m-N}{N} +1}\right)^{\frac{N}{m-N}} \\ &\to 1 \end{split} $$

as $m\to \infty$. While the other term tends to $0$. So

$$\left(\frac N m \right)^{\frac{N}{m-N}}-\left(\frac N m \right)^{\frac{m}{m-N}} \to 1$$

as $m\to \infty$.

Answer 2

A uniformly Cauchy sequence is pointwise convergent, and it uniformly converges to that pointwise limit.

The pointwise limit of the $\{f_n\}$ is $0$ on $(0,1)$ and $1$ at $1$. Since a uniform limit of continuous functions is continuous, this is a contradiction. Your sequence is not uniformly Cauchy.

Answer 3

It suffices to show there is some $\epsilon_0 >0$ such that for every $n$ there is some $m > n$ with $\lVert f_m - f_n \rVert > \epsilon_0$.

Choose $\epsilon_0 = 1/4$ and $m = 2n$.

Note that for $x_n = (1/2)^{1/n}$ we have $\lVert f_m-f_n \rVert \geqslant |x_n^{2n}-x_n^{n}| = 1/4$.